Question

The sum of $n$ terms of the following series $1+(1+x)+(1+x+x^2)+...$will be

• A. $\frac{(1-x^n)}{(1-x)}$
• B. $\frac{x(1-x^n)}{(1-x)}$
• C. $\frac{\left[n\right(1-x)–x(1-x^n\left)\right]}{{(1-x)}^2}$
• D. None of these

Answer 1

The correct option is C

$\frac{\left[n\right(1-x)–x(1-x^n\left)\right]}{{(1-x)}^2}$

Explanation for the correct option:

Find the required sum.

Given: series$1+(1+x)+(1+x+x^2)+...$$n$ terms

Let $S=1+(1+x)+(1+x+x^2)+\dots n$ terms …(i)

$xS=x+(x+x^2)+(x+x^2+x^3)+\dots n$ terms ..(ii)

Subtracting (ii) from (i), we get

$(1-x)S=n–x\frac{(1-x^n)}{(1-x)}\dots ..$( Sum of GP series)

$$\begin{gathered} \Rightarrow S=\frac{n}{(1-x)}–x\frac{(1-x^n)}{(1-x{)}^2} \\ \Rightarrow S=\frac{\left[n\right(1-x)–x(1-x^n\left)\right]}{{(1-x)}^2} \end{gathered}$$

Hence, option (C) is the correct answer.